Question

Given $K_{SP}(Ag_2C_2O_4) = 32X$, $K_{SP}(AgBr) = 4Y$. Find ratio of solubility of the given salts in pure water

• A. $\frac{X^{1/3}}{\sqrt{2Y}}$
• B. $\frac{2X^{1/3}}{\sqrt{2Y}}$
• C. $\frac{2X^{1/3}}{\sqrt{Y}}$
• D. $\frac{X^{1/3}}{\sqrt{Y}}$

Answer 1

The Correct Option is D

The solubility product constant ($K_{SP}$) relates to the solubility ($S$) of a salt based on its dissociation stoichiometry.
1. For Silver Oxalate ($Ag_2C_2O_4$):
Dissociation: $Ag_2C_2O_4(s) \rightleftharpoons 2Ag^+(aq) + C_2O_4^{2-}(aq)$
If solubility is $S_1$, then $[Ag^+] = 2S_1$ and $[C_2O_4^{2-}] = S_1$.
$K_{SP} = [Ag^+][C_2O_4^{2-}] = (2S_1)^2 \cdot S_1 = 4S_1^3$
Given $K_{SP} = 32X$:
$4S_1^3 = 32X \Rightarrow S_1^3 = 8X \Rightarrow S_1 = 2X^{1/3}$

2. For Silver Bromide ($AgBr$):
Dissociation: $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$
If solubility is $S_2$, then $[Ag^+] = S_2$ and $[Br^-] = S_2$.
$K_{SP} = [Ag^+][Br^-] = S_2^2$
Given $K_{SP} = 4Y$:
$S_2^2 = 4Y \Rightarrow S_2 = 2Y^{1/2}$

3. Calculate the ratio:
$\text{Ratio} = \frac{S_1}{S_2} = \frac{2X^{1/3}}{2Y^{1/2}} = \frac{X^{1/3}}{\sqrt{Y}}$

This matches option (4).